(a) Here, potential difference across the bulb, $\mathrm{V}=200\mathrm{V}$
Power of the bulb, $\mathrm{P}=100\mathrm{W}$
As $\mathrm{P}=\frac{{\mathrm{V}}^2}{\mathrm{R}},\mathrm{R}=\frac{{\mathrm{V}}^2}{\mathrm{P}}=\frac{{\left(200\mathrm{V}\right)}^2}{100\mathrm{W}}$
$=\frac{4\times {10}^4}{100}\mathit{\Omega }=400\mathit{\Omega }$
(b) Electric energy consumed by 1 bulb in 10 hours for 30 days, i.e.,
$\mathrm{W}=\mathrm{P}\mathit{t}$
$=100\mathrm{W}\times 10\times 30$
$=30000\mathrm{Wh}$
Electric energy consumed by 3 bulbs
$=30000\times 3$
$=90000\mathrm{Wh}$
$=90\mathrm{kWh}$

(c) Cost of 90 units of electric energy

OR
(a) Ohm's law states that current flowing through a conductor is directly proportional to the potential difference maintained across the two ends of a conductor at constant temperature and pressure.
Let current flowing through a conductor is $\mathrm{I},\mathrm{V}$ is the potential difference maintained across the two ends of conductor.
As per Ohm's law, $\mathrm{V}\propto \mathrm{I}$
Thus,
$\mathrm{V}=\mathit{I}\mathit{R}$
Here $\mathrm{R}$ is a proportionality constant called resistance.
As per the question, the resistance of a conductor is $1\mathrm{ohm}$.
Let $\mathit{V}=1$ volt and $\mathit{I}=1\mathrm{A}$, then $\mathit{R}=1\mathrm{ohm}$ Hence, the resistance is said to be $1\mathrm{ohm}$ if 1 ampere of current flows through a circuit due to the potential difference of 1 volt.
(b) Electric power is the rate at which work is done or energy is transformed in an electrical circuit.
The formula for electric power is given by:
$\mathit{P}=\mathit{V}\mathit{I}$
where,
$\mathrm{P}$ is the power
$\mathrm{V}$ is the potential difference in the circuit I is the electric current
Power can also be written as
$\mathit{P}={\mathit{I}}^2\mathit{R}$
$\mathit{P}=\frac{{\mathit{V}}^2}{\mathit{R}}$
The above two expressions are got by using Ohms law, where, voltage, current
and resistance are related by the following relation.
Where,
$\mathit{R}$ is the resistance in the circuit.
$\mathrm{V}$ is the potential difference in the circuit I is the electric current
(c) For $\mathit{x}$ number of resistors of resistance $132\mathit{\Omega }$
Supply voltage, $\mathrm{V}=220\mathrm{V}$
Current, $\mathrm{I}=5\mathrm{A}$
Equivalent resistance of the combination $=\mathit{R}$, given as
$1∕\mathrm{R}=\mathit{x}\times \left(1∕132\right)$
$\mathrm{R}=132∕\mathit{x}$
From Ohm's law :
$132\mathit{x}=220∕5$
$132\mathit{x}=44$
$132=44\mathit{x}$
$\mathit{x}=132∕44$
$\mathit{x}=3$
So, answer is 3 resistors.